I have a problem where I am asked to find \$i_L(t)\$ and \$v_c(t)\$ for a parallel RLC circuit, with a step voltage source.

The circuit diagram is this below, where the voltage source is \$5 u(t)\$, or \$V=0, t<0\$ and \$V=5, t\ge0\$.:

circuit diagram image in pspice

So far, I found \$\alpha=\dfrac{1}{2RC}=3.33 *10^5\$ rad/s, and \$\omega_o=\dfrac{1}{\sqrt{LC}}=3.16*10^5\$ rad/s.

Since \$\alpha^2 > \omega_0^2\$, it is overdamped, and thus the general solution of the DE has the form \$i_L(t)=Ae^{s_1t}+Be^{s_2t}+i_L(\infty)\$.

The roots of the characteristic equation are, \$S_1\$ and \$S_2\$ are \$ \{-2.279*10^5, -4.387*10^5\} \$.

It is known that:

\$ i_L(0^-) = 0 \$ A, since there is no source at that time.

\$ i_L(\infty) = 0.033 \$ A, since capacitor would be fully charged (no current flows), and the inductor would essentially act as a short.

The part I'm having trouble is determining what \$i_L(0^+)\$ would be. In this circuit, how would the inductor current be determined immediately after \$t = 0\$?


2 Answers 2


The current through an (ideal) inductor is continuous, just as the electric potential difference over a capacitor. So if the current \$ i_L \$ at \$t=0-\$ is zero, then so is the current at \$t=0+\$, immediately after the voltage source stepped to \$+5 V\$.

If you want a full solution of capacitor voltage and inductor current, you'd have to solve the differential equation for e.g. the capacitor voltage \$ u_C\$. $$ u_S(t) = 5 * UnitStep $$ $$ i_R(t) = (u_S(t) - u_C(t))/R$$ $$ i_L(t) = \frac{1}{L}\int_0^t u_C(\tau) d\tau $$ $$ i_C(t) = C \frac{du_C}{dt} $$

In which the subscripts are quite self-explaining, except \$u_S(t)\$, which is the source voltage.

Now, omitting the time where possible, we can write $$u_c = u_S - R(i_L + i_C) = u_S - R\left(\frac{1}{L}\int u_C dt +C\frac{du_C}{dt}\right) $$ After differentiating the whole expression we get, and we write u for u_C, $$ u'= -\frac{R}{L}u-RCu'' $$ or $$ RCu''+u'+\frac{R}{L}u = 0 $$ Substituting the values for \$R\$, \$L\$ and \$C\$ gives $$ 150*10^{-8}u''+u'+ 150*10^3u = 0 $$ Multiplying with \$2*10^6\$ gives us $$ 3u''(t) + 2 * 10^6 u'(t) + 3 * 10^{11} u(t) = 0 $$

Some boundary conditions of this problem are \$u(0)\$ and \$u'(0)\$. Of course \$u(0)=0\$ as at the start the capacitor is empty. \$u'(0)\$ can be found by realizing ourselves that \$i_L(0) = 0\$ because the current in any self inductance is continuous and therefore all the current at \$t=0\$ will flow into the capacitor. This current is caused by the voltage drop of \$5 V\$ over the resistors when the step is enacted and C has zero charge hence zero voltage. So this current is \$ i_C(0+) = 5/150 A \$ which causes a rate of change of the capacitor voltage \$ u'_C(0+) = C*i_C(0+) = \frac{5}{150}*10^{-8} V/s\$. Therefore the boundary conditions are $$ u(0) = 0 $$ $$ u'(0) = \frac{1}{3} * 10^7 $$

The exact solution of this differential equation with boundary conditions is

$$ u_C(t) = 5 * \sqrt 10 * e^{-10^5 (10 + \sqrt 10) t/3} (e^{2*10^5 \sqrt(10) t/3} - 1) $$ and it plots like this.

Now this deals with the voltage \$u_C(t)\$ over the capacitor and your question was regarding the inductor current which is $$ i_L(t) = \frac{1}{L} \int_0^t u(\tau) d\tau $$ and with \$a=\sqrt 10\$, \$b=100000/3\$ (and \$a^2=10\$) this becomes $$i_L(t) = \frac{1 - e^{-a^2 b t} \left(cosh(a b t) + a sinh(a b t)\right)}{b(1 - 1/a^2)} $$

Of course, when \$t=0\$, the exponent becomes \$e^{-0}=1\$, \$sinh(0)=0\$ and \$cosh(0)=1\$, hence the whole expression becomes $$ i_L(0+) = \frac{1-1}{b(1-1/a^2)}=0 $$


The part I'm having trouble is determining what iL(0+) would be. In this circuit, how would the inductor current be determined immediately after t=0?

As with any inductor, the current cannot change instantly so, the current immediately after t = 0 is zero.


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